Re: [Vo]:Re: centripetal force question
On 25/7/2007 2:01 PM, Harry Veeder wrote: > On 21/7/2007 11:01 PM, Harry Veeder wrote: > >> Something else leaves me wondering... >> >>> Harry wrote: >>> While the brakes are applied the wheel is not turning about its own centre. >> >> Michel wrote: >>> Wrt the ground it isn't, but wrt the distant stars it is, at the >>> rate of one turn per day (it can't be non-rotating wrt both, >>> agreed?). >> >> >> If you began to ride the wheel nearest the ground after the brake is >> realeased the net centrifugal force you would feel should depend >> only on the rotation of the earth, since the wheel is not rotating >> wrt to the ground. >> >> However, wrt to the distant stars the wheel is rotating so you should >> feel a reduced centrifugal force. >> >> Apparently theory leads to a contradiction. >> >> Or is my reasoning faulty? >> >> Harry >> > > How does this sound? > > The centrifugal force on the person due to the Earth's rotation with the > brake on is inversely proportional to radius of the Earth. oops, that should be directly proportional to the radius of the Earth, (if the earth's rotation is constant). > However, when the brake is off the centrifugal force is inversely > proportional to the radius of the Earth PLUS the radius of the wheel. also directly proportional to the radius of Earth plus radius of wheel. > Therefore, although the path traced by the person is the same in both > cases the net centrifugal force is different. > > It would seem the way you are connected to the Earth can affect your > apparent weight. However, this dependence would disappear if the Earth were > not rotating. > > Harry > Harry
Re: [Vo]:Re: centripetal force question
On 21/7/2007 11:01 PM, Harry Veeder wrote: > Something else leaves me wondering... > >> Harry wrote: >> >>> While the brakes are applied the wheel is not turning about >>> its own centre. > > Michel wrote: >> Wrt the ground it isn't, but wrt the distant stars it is, at the >> rate of one turn per day (it can't be non-rotating wrt both, >> agreed?). > > > If you began to ride the wheel nearest the ground after the brake is > realeased the net centrifugal force you would feel should depend > only on the rotation of the earth, since the wheel is not rotating > wrt to the ground. > > However, wrt to the distant stars the wheel is rotating so you should > feel a reduced centrifugal force. > > Apparently theory leads to a contradiction. > > Or is my reasoning faulty? > > Harry > How does this sound? The centrifugal force on the person due to the Earth's rotation with the brake on is inversely proportional to radius of the Earth. However, when the brake is off the centrifugal force is inversely proportional to the radius of the Earth PLUS the radius of the wheel. Therefore, although the path traced by the person is the same in both cases the net centrifugal force is different. It would seem the way you are connected to the Earth can affect your apparent weight. However, this dependence would disappear if the Earth were not rotating. Harry
Re: [Vo]:Re: centripetal force question
On 21/7/2007 2:52 PM, Horace Heffner wrote: On Jul 21, 2007, at 10:55 AM, Harry Veeder wrote: But if the hand of God placed a non-rotating wheel (wrt to distant stars) in the supporting frame, then it would behave as I drew it? ;-) All you would need to get the rotation right would be a sidereal drive from a telescope equatorial mount. If you extracted any energy from the wheel motion wrt earth though, its angular velocity would change to closer to that of earth, until it matched and no more ennergy would be available. No free lunch there, but the answer to the above question is yes. Yes, although in this case I wasn't specifically looking for a free lunch... but maybe a free ride. ;-) Harry
Re: [Vo]:Re: centripetal force question
On 22/7/2007 4:30 PM, Michel Jullian wrote: > > - Original Message - > From: "Harry Veeder" <[EMAIL PROTECTED]> > To: > Sent: Sunday, July 22, 2007 10:33 PM > Subject: Re: [Vo]:Re: centripetal force question > > > ... >>> Interesting thread. The reason I started it was because I'd like >>> to >>> know, suppose the earth started spinning faster, say 10 times it's >>> present rotational speed, how much weight would I loose? >>> >> >> >> your weight lost in newtons would be: mr(w^2) >> >> where m is your mass in kg, >> w is revolutions/second, > > radians/second actually (2*pi*rotations/second) > yes, indeed. Harry >> and r is the radius of earth in meters. >> >> To convert newtons to pounds multiply by .225 >
[Vo]:Re: centripetal force question
- Original Message - From: "Harry Veeder" <[EMAIL PROTECTED]> To: Sent: Sunday, July 22, 2007 10:33 PM Subject: Re: [Vo]:Re: centripetal force question ... >> Interesting thread. The reason I started it was because I'd like >> to >> know, suppose the earth started spinning faster, say 10 times it's >> present rotational speed, how much weight would I loose? >> > > > your weight lost in newtons would be: mr(w^2) > > where m is your mass in kg, > w is revolutions/second, radians/second actually (2*pi*rotations/second) > and r is the radius of earth in meters. > > To convert newtons to pounds multiply by .225
Re: [Vo]:Re: centripetal force question
- Original Message - From: thomas malloy <[EMAIL PROTECTED]> Date: Sunday, July 22, 2007 1:36 am Subject: Re: [Vo]:Re: centripetal force question > Harry Veeder wrote: > > > > Interesting thread. The reason I started it was because I'd like > to > know, suppose the earth started spinning faster, say 10 times it's > present rotational speed, how much weight would I loose? > your weight lost in newtons would be: mr(w^2)where m is your mass in kg, w is revolutions/second, and r is the radius of earth in meters. To convert newtons to pounds multiply by .225 Harry
Re: [Vo]:Re: centripetal force question
Harry Veeder wrote: Interesting thread. The reason I started it was because I'd like to know, suppose the earth started spinning faster, say 10 times it's present rotational speed, how much weight would I loose? --- http://USFamily.Net/dialup.html - $8.25/mo! -- http://www.usfamily.net/dsl.html - $19.99/mo! ---
Re: [Vo]:Re: centripetal force question
Something else leaves me wondering... >Harry wrote: > >> While the brakes are applied the wheel is not turning about >> its own centre. Michel wrote: > Wrt the ground it isn't, but wrt the distant stars it is, at the > rate of one turn per day (it can't be non-rotating wrt both, > agreed?). If you began to ride the wheel nearest the ground after the brake is realeased the net centrifugal force you would feel should depend only on the rotation of the earth, since the wheel is not rotating wrt to the ground. However, wrt to the distant stars the wheel is rotating so you should feel a reduced centrifugal force. Apparently theory leads to a contradiction. Or is my reasoning faulty? Harry
Re: [Vo]:Re: centripetal force question
On Jul 21, 2007, at 10:55 AM, Harry Veeder wrote: But if the hand of God placed a non-rotating wheel (wrt to distant stars) in the supporting frame, then it would behave as I drew it? ;-) All you would need to get the rotation right would be a sidereal drive from a telescope equatorial mount. If you extracted any energy from the wheel motion wrt earth though, its angular velocity would change to closer to that of earth, until it matched and no more ennergy would be available. No free lunch there, but the answer to the above question is yes. Horace Heffner http://www.mtaonline.net/~hheffner/
[Vo]:Re: centripetal force question
Certainly, even the Hand Of Harry could do it, by setting it in rotation wrt the ground at one turn per day and then releasing it. Michel - Original Message - From: "Harry Veeder" <[EMAIL PROTECTED]> To: Sent: Saturday, July 21, 2007 8:55 PM Subject: Re: [Vo]:Re: centripetal force question ... > But if the hand of God placed a non-rotating wheel (wrt to distant stars) > > in the supporting frame, then it would behave as I drew it? ;-) ...
Re: [Vo]:Re: centripetal force question
- Original Message - From: Michel Jullian <[EMAIL PROTECTED]> Date: Friday, July 20, 2007 8:38 pm Subject: [Vo]:Re: centripetal force question > Harry wrote: > > > While the brakes are applied the wheel is not turning about its > own centre. > > Wrt the ground it isn't, but wrt the distant stars it is, at the > rate of one turn per day (it can't be non-rotating wrt both, > agreed?). As I said, and as Steven paraphrased and Horace finally > opined: > "...If it's initially non-rotating wrt the Earth, i.e. rotating > once per day wrt the distant stars, it will keep its angular speed > i.e. it will remain at rest wrt the Earth." > > IOW what you drew (wheel not rotating wrt the distant stars) is > not what will happen when the brakes are released, on the contrary > the wheel will keep rotating wrt the stars at the sa! me rate a But if the hand of God placed a non-rotating wheel (wrt to distant stars) in the supporting frame, then it would behave as I drew it? ;-) Harry
RE: [Vo]:Re: centripetal force question
The term gyration is given as a synonym for rotation sometimes, but I think of gyration and rotation as different things. I think gyration is just something following a circular path in at least 2 dimensions and is in fact a form of linear motion. True rotation happens when something shows a distinguishable "face" or a repeating series of faces of some sort to a fixed reference point in a cyclical manner as it goes about "rotating", and can occur in just one dimension. The true deep meaning of rotation escapes me (spin-2, etc.), because the concept is difficult and I'm not always sure what does or doesn't constitute a face. This all gets more interesting when the rotating(?) gyrating(?) things are charged and the fixed frame they are in has a magnetic field (or vice-versa). Long threads on Vort about whether the fields actually rotate with their sources, etc. Rick M.
[Vo]:Re: centripetal force question
Harry wrote: > While the brakes are applied the wheel is not turning about its own centre. Wrt the ground it isn't, but wrt the distant stars it is, at the rate of one turn per day (it can't be non-rotating wrt both, agreed?). As I said, and as Steven paraphrased and Horace finally opined: "...If it's initially non-rotating wrt the Earth, i.e. rotating once per day wrt the distant stars, it will keep its angular speed i.e. it will remain at rest wrt the Earth." IOW what you drew (wheel not rotating wrt the distant stars) is not what will happen when the brakes are released, on the contrary the wheel will keep rotating wrt the stars at the same rate as before, and therefore will remain immobile wrt the ground, unfortunately (if it had worked as you drew it would have been a nice way to produce energy, by attaching a dynamo to the wheel!). Michel
Re: [Vo]:Re: centripetal force question
On 20/7/2007 1:50 PM, OrionWorks wrote: > On 7/20/07, Harry Veeder wrote: >> I made a drawing of the situation I was imagining: >> >> http://web.ncf.ca/eo200/world-ferriswheel.html >> >> A ferris wheel located on Earth's equator. Initially a brake prevents the >> wheel from turning. After the brake is released and assuming the axel of the >> wheel is frictionless, will the orientation of the wheel remain unchanged as >> the Earth revolves? >> >> Harry >> > > Harry, > > Nice Illustration. What graphics package do you use? Thanks. I use Adobe Illustrator. > I should think that the ferris wheel will appear stationary to a > ground observer - after the break has been released. IOW, from a > so-called gods-eye view, away from the planet's surface, the ferris > wheel will be seen to be rotating at the same RPM speed as the planet. > The RPM bestowed on the ferris wheel via inertia once the breaks have > been release will not have changed, therefore the wheel will appear to > rotate at the same RPM speed as the Earth as perceived off of the > planet, but will appear stationary to a ground observer. > > Does anyone wish to disagree? Did I miss something? While the brakes are applied the wheel is not turning about its own centre. Harry
Re: [Vo]:Re: centripetal force question
Just nit pcking, but I'd disagree, It may even spin in the opposite direction if the wheel reaches the jet stream at the right point. Not being too serious here, who is to say if from the God perspective the ferris wheel appears to turn or the earth appears to turn. However, from the ground since YOUR stationary point of view is from the earth, then the ferris wheel would appear to turn, not the other way around. ;^) G --- OrionWorks <[EMAIL PROTECTED]> wrote: > On 7/20/07, Harry Veeder wrote: > > I made a drawing of the situation I was imagining: > > > > http://web.ncf.ca/eo200/world-ferriswheel.html > > > > A ferris wheel located on Earth's equator. > Initially a brake prevents the > > wheel from turning. After the brake is released > and assuming the axel of the > > wheel is frictionless, will the orientation of the > wheel remain unchanged as > > the Earth revolves? > > > > Harry > > > > Harry, > > Nice Illustration. What graphics package do you use? > > I should think that the ferris wheel will appear > stationary to a > ground observer - after the break has been released. > IOW, from a > so-called gods-eye view, away from the planet's > surface, the ferris > wheel will be seen to be rotating at the same RPM > speed as the planet. > The RPM bestowed on the ferris wheel via inertia > once the breaks have > been release will not have changed, therefore the > wheel will appear to > rotate at the same RPM speed as the Earth as > perceived off of the > planet, but will appear stationary to a ground > observer. > > Does anyone wish to disagree? Did I miss something? > > Regards, > Steven Vincent Johnson > www.OrionWorks.com > > Luggage? GPS? Comic books? Check out fitting gifts for grads at Yahoo! Search http://search.yahoo.com/search?fr=oni_on_mail&p=graduation+gifts&cs=bz
Re: [Vo]:Re: centripetal force question
why... wouldnt there be? the wheel WILL turn, that much is certain (turn reltive to the fixed point on the earth, that is ohh, i see what you mean. its the wheels own angular momentum being conserved, not any from the earth being transferred to it, right? On 7/20/07, Harry Veeder <[EMAIL PROTECTED]> wrote: On 20/7/2007 2:48 PM, leaking pen wrote: > on a more serious answer, if you are only taking the earths rotation > into account, then yes, i would think so. remember, becuase of the > tilt of teh earth, 90 degrees straight with teh ground would not be > the right orientation, me thinks. I'm ignoring the tilt of the Earth relative to the orbit it makes around the sun. > How do you measure the force involved in a transfer of angular > momentum like that? I'm not sure there is a _transfer_ of angular momentum...which is why I find this "thought experiment" intriguing. Harry > On 7/20/07, Harry Veeder <[EMAIL PROTECTED]> wrote: >> I made a drawing of the situation I was imagining: >> >> http://web.ncf.ca/eo200/world-ferriswheel.html >> >> A ferris wheel located on Earth's equator. Initially a brake prevents the >> wheel from turning. After the brake is released and assuming the axel of the >> wheel is frictionless, will the orientation of the wheel remain unchanged as >> the Earth revolves? >> >> Harry >> >> >> On 20/7/2007 5:42 AM, Michel Jullian wrote: >> >>> Lets' assume the wheel axis is parallel to the Earth's for simplicity. >>> Moving >>> the hub around in the plane of the wheel shouldn't change its rate of >>> rotation >>> wrt an inertial frame if there is no friction. >>> >>> If it's initially non-rotating wrt the distant stars (i.e. rotating once per >>> day wrt the Earth) it will remain so, you can verify that with a bicycle, >>> lift >>> the front you'll see that the front wheel keeps the same orientation. >>> >>> If it's initially non-rotating wrt the Earth, i.e. rotating once per day wrt >>> the distant stars, it will keep its angular speed i.e. it will remain at >>> rest >>> wrt the Earth. >>> >>> Michel >>> >>> - Original Message - >>> From: "Harry Veeder" <[EMAIL PROTECTED]> >>> To: >>> Sent: Friday, July 20, 2007 5:23 AM >>> Subject: Re: [Vo]:centripetal force question >>> >>> Imagine a ferris wheel (absent the carriages for simplicity) which is initially at rest. As the world turns, the ferris wheel will complete one revolution in a day (assuming no friction) with respect to an observer standing beside it. Yes? No? ...or? Harry On 15/7/2007 5:17 PM, Michel Jullian wrote: > Hi Thomas, > > The (fictitious, or apparent) force you're talking about is a function of > _rotations_ (not revolutions) per second, and also of your mass and of > your > distance from the axis > (force=mass*(2*pi*rotations_per_second)^2*radius_of_the_earth assuming you > stand on the equator), nothing to do with distance traveled by the planet, > and > it is not centripetal (going towards the center) but centrifugal (think of > fugitive = going away from the center), if it was centripetal it would not > subtract from but add to actual weight, which is the actual centripetal > force. > > Michel > > - Original Message - > From: "thomas malloy" <[EMAIL PROTECTED]> > To: > Sent: Sunday, July 15, 2007 10:02 PM > Subject: [Vo]:centripital force question > > >> I'm subject to weight loss produced by the centripital force produced by >> the earth's rotation, I'm wondering if centripetal force is a function >> of revolutions per time unit, or total distance traveled as the planet >> travels? >> >> >> --- http://USFamily.Net/dialup.html - $8.25/mo! -- >> http://www.usfamily.net/dsl.html - $19.99/mo! --- >> > >>> >> >> > -- That which yields isn't always weak.
Re: [Vo]:Re: centripetal force question
On 20/7/2007 2:48 PM, leaking pen wrote: > on a more serious answer, if you are only taking the earths rotation > into account, then yes, i would think so. remember, becuase of the > tilt of teh earth, 90 degrees straight with teh ground would not be > the right orientation, me thinks. I'm ignoring the tilt of the Earth relative to the orbit it makes around the sun. > How do you measure the force involved in a transfer of angular > momentum like that? I'm not sure there is a _transfer_ of angular momentum...which is why I find this "thought experiment" intriguing. Harry > On 7/20/07, Harry Veeder <[EMAIL PROTECTED]> wrote: >> I made a drawing of the situation I was imagining: >> >> http://web.ncf.ca/eo200/world-ferriswheel.html >> >> A ferris wheel located on Earth's equator. Initially a brake prevents the >> wheel from turning. After the brake is released and assuming the axel of the >> wheel is frictionless, will the orientation of the wheel remain unchanged as >> the Earth revolves? >> >> Harry >> >> >> On 20/7/2007 5:42 AM, Michel Jullian wrote: >> >>> Lets' assume the wheel axis is parallel to the Earth's for simplicity. >>> Moving >>> the hub around in the plane of the wheel shouldn't change its rate of >>> rotation >>> wrt an inertial frame if there is no friction. >>> >>> If it's initially non-rotating wrt the distant stars (i.e. rotating once per >>> day wrt the Earth) it will remain so, you can verify that with a bicycle, >>> lift >>> the front you'll see that the front wheel keeps the same orientation. >>> >>> If it's initially non-rotating wrt the Earth, i.e. rotating once per day wrt >>> the distant stars, it will keep its angular speed i.e. it will remain at >>> rest >>> wrt the Earth. >>> >>> Michel >>> >>> - Original Message - >>> From: "Harry Veeder" <[EMAIL PROTECTED]> >>> To: >>> Sent: Friday, July 20, 2007 5:23 AM >>> Subject: Re: [Vo]:centripetal force question >>> >>> Imagine a ferris wheel (absent the carriages for simplicity) which is initially at rest. As the world turns, the ferris wheel will complete one revolution in a day (assuming no friction) with respect to an observer standing beside it. Yes? No? ...or? Harry On 15/7/2007 5:17 PM, Michel Jullian wrote: > Hi Thomas, > > The (fictitious, or apparent) force you're talking about is a function of > _rotations_ (not revolutions) per second, and also of your mass and of > your > distance from the axis > (force=mass*(2*pi*rotations_per_second)^2*radius_of_the_earth assuming you > stand on the equator), nothing to do with distance traveled by the planet, > and > it is not centripetal (going towards the center) but centrifugal (think of > fugitive = going away from the center), if it was centripetal it would not > subtract from but add to actual weight, which is the actual centripetal > force. > > Michel > > - Original Message - > From: "thomas malloy" <[EMAIL PROTECTED]> > To: > Sent: Sunday, July 15, 2007 10:02 PM > Subject: [Vo]:centripital force question > > >> I'm subject to weight loss produced by the centripital force produced by >> the earth's rotation, I'm wondering if centripetal force is a function >> of revolutions per time unit, or total distance traveled as the planet >> travels? >> >> >> --- http://USFamily.Net/dialup.html - $8.25/mo! -- >> http://www.usfamily.net/dsl.html - $19.99/mo! --- >> > >>> >> >> >
Re: [Vo]:Re: centripetal force question
Ya'll should check the rest of that comic out, ive fallen in love with it. excellent math humor throughout. On 7/20/07, Jed Rothwell <[EMAIL PROTECTED]> wrote: Horace Heffner wrote: The spelling mistake must have have been due to my lack of attention from laughing so hard at the cartoon leaking pen posted: http://www.xkcd.com/c123.html That is funny. - Jed -- That which yields isn't always weak.
Re: [Vo]:Re: centripetal force question
Horace Heffner wrote: The spelling mistake must have have been due to my lack of attention from laughing so hard at the cartoon leaking pen posted: http://www.xkcd.com/c123.html That is funny. - Jed
Re: [Vo]:Re: centripetal force question
On Jul 20, 2007, at 11:21 AM, Jed Rothwell wrote: Horace Heffner wrote: Sounds right to me. I overlooked the break aspect. When the breaks are on the wheel is actually rotating . . . You mean BRAKE. We are talking about a brake on the wheel, not a break on a broken wheel. You had me confused there. Sorry about that! The spelling mistake must have have been due to my lack of attention from laughing so hard at the cartoon leaking pen posted: http://www.xkcd.com/c123.html Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Re: centripetal force question
on a more serious answer, if you are only taking the earths rotation into account, then yes, i would think so. remember, becuase of the tilt of teh earth, 90 degrees straight with teh ground would not be the right orientation, me thinks. How do you measure the force involved in a transfer of angular momentum like that? On 7/20/07, Harry Veeder <[EMAIL PROTECTED]> wrote: I made a drawing of the situation I was imagining: http://web.ncf.ca/eo200/world-ferriswheel.html A ferris wheel located on Earth's equator. Initially a brake prevents the wheel from turning. After the brake is released and assuming the axel of the wheel is frictionless, will the orientation of the wheel remain unchanged as the Earth revolves? Harry On 20/7/2007 5:42 AM, Michel Jullian wrote: > Lets' assume the wheel axis is parallel to the Earth's for simplicity. Moving > the hub around in the plane of the wheel shouldn't change its rate of rotation > wrt an inertial frame if there is no friction. > > If it's initially non-rotating wrt the distant stars (i.e. rotating once per > day wrt the Earth) it will remain so, you can verify that with a bicycle, lift > the front you'll see that the front wheel keeps the same orientation. > > If it's initially non-rotating wrt the Earth, i.e. rotating once per day wrt > the distant stars, it will keep its angular speed i.e. it will remain at rest > wrt the Earth. > > Michel > > - Original Message - > From: "Harry Veeder" <[EMAIL PROTECTED]> > To: > Sent: Friday, July 20, 2007 5:23 AM > Subject: Re: [Vo]:centripetal force question > > >> >> Imagine a ferris wheel (absent the carriages for simplicity) which is >> initially at rest. >> >> As the world turns, the ferris wheel will complete one revolution in a day >> (assuming no friction) with respect to an observer standing >> beside it. >> >> Yes? No? ...or? >> >> Harry >> >> >> >> >> On 15/7/2007 5:17 PM, Michel Jullian wrote: >> >>> Hi Thomas, >>> >>> The (fictitious, or apparent) force you're talking about is a function of >>> _rotations_ (not revolutions) per second, and also of your mass and of your >>> distance from the axis >>> (force=mass*(2*pi*rotations_per_second)^2*radius_of_the_earth assuming you >>> stand on the equator), nothing to do with distance traveled by the planet, >>> and >>> it is not centripetal (going towards the center) but centrifugal (think of >>> fugitive = going away from the center), if it was centripetal it would not >>> subtract from but add to actual weight, which is the actual centripetal >>> force. >>> >>> Michel >>> >>> - Original Message - >>> From: "thomas malloy" <[EMAIL PROTECTED]> >>> To: >>> Sent: Sunday, July 15, 2007 10:02 PM >>> Subject: [Vo]:centripital force question >>> >>> I'm subject to weight loss produced by the centripital force produced by the earth's rotation, I'm wondering if centripetal force is a function of revolutions per time unit, or total distance traveled as the planet travels? --- http://USFamily.Net/dialup.html - $8.25/mo! -- http://www.usfamily.net/dsl.html - $19.99/mo! --- >>> >> > -- That which yields isn't always weak.
Re: [Vo]:Re: centripetal force question
Horace Heffner wrote: Sounds right to me. I overlooked the break aspect. When the breaks are on the wheel is actually rotating . . . You mean BRAKE. We are talking about a brake on the wheel, not a break on a broken wheel. You had me confused there. - Jed
Re: [Vo]:Re: centripetal force question
On Jul 20, 2007, at 10:50 AM, OrionWorks wrote: I should think that the ferris wheel will appear stationary to a ground observer - after the break has been released. IOW, from a so-called gods-eye view, away from the planet's surface, the ferris wheel will be seen to be rotating at the same RPM speed as the planet. The RPM bestowed on the ferris wheel via inertia once the breaks have been release will not have changed, therefore the wheel will appear to rotate at the same RPM speed as the Earth as perceived off of the planet, but will appear stationary to a ground observer. Sounds right to me. I overlooked the break aspect. When the breaks are on the wheel is actually rotating 1 rev/day, which won't stop when the breaks are released because the bearings etc. are assumed to be frictionless. Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Re: centripetal force question
On 7/20/07, Harry Veeder wrote: I made a drawing of the situation I was imagining: http://web.ncf.ca/eo200/world-ferriswheel.html A ferris wheel located on Earth's equator. Initially a brake prevents the wheel from turning. After the brake is released and assuming the axel of the wheel is frictionless, will the orientation of the wheel remain unchanged as the Earth revolves? Harry Harry, Nice Illustration. What graphics package do you use? I should think that the ferris wheel will appear stationary to a ground observer - after the break has been released. IOW, from a so-called gods-eye view, away from the planet's surface, the ferris wheel will be seen to be rotating at the same RPM speed as the planet. The RPM bestowed on the ferris wheel via inertia once the breaks have been release will not have changed, therefore the wheel will appear to rotate at the same RPM speed as the Earth as perceived off of the planet, but will appear stationary to a ground observer. Does anyone wish to disagree? Did I miss something? Regards, Steven Vincent Johnson www.OrionWorks.com
Re: [Vo]:Re: centripetal force question
On Jul 20, 2007, at 11:33 AM, Harry Veeder wrote: I made a drawing of the situation I was imagining: http://web.ncf.ca/eo200/world-ferriswheel.html This might work if you had frictionless bearings. A better design to illustrate the principle is the Foucault pendulum: http://en.wikipedia.org/wiki/Foucault_pendulum They make for neat clocks. I have actually seen them in action. They do not keep track of 24 hour time though, they track the sidereal time, which is relative to the fixed stars and not the sun. An ideal device would complete a rotation in a sidereal day: 23 hours, 56 minutes, 4.091 seconds. Horace Heffner http://www.mtaonline.net/~hheffner/
Re: [Vo]:Re: centripetal force question
http://www.xkcd.com/c123.html that is all i have to say on teh subject On 7/20/07, Horace Heffner <[EMAIL PROTECTED]> wrote: On Jul 20, 2007, at 11:33 AM, Harry Veeder wrote: > I made a drawing of the situation I was imagining: > > http://web.ncf.ca/eo200/world-ferriswheel.html This might work if you had frictionless bearings. A better design to illustrate the principle is the Foucault pendulum: http://en.wikipedia.org/wiki/Foucault_pendulum They make for neat clocks. I have actually seen them in action. They do not keep track of 24 hour time though, they track the sidereal time, which is relative to the fixed stars and not the sun. An ideal device would complete a rotation in a sidereal day: 23 hours, 56 minutes, 4.091 seconds. Horace Heffner http://www.mtaonline.net/~hheffner/ -- That which yields isn't always weak.
Re: [Vo]:Re: centripetal force question
I made a drawing of the situation I was imagining: http://web.ncf.ca/eo200/world-ferriswheel.html A ferris wheel located on Earth's equator. Initially a brake prevents the wheel from turning. After the brake is released and assuming the axel of the wheel is frictionless, will the orientation of the wheel remain unchanged as the Earth revolves? Harry On 20/7/2007 5:42 AM, Michel Jullian wrote: > Lets' assume the wheel axis is parallel to the Earth's for simplicity. Moving > the hub around in the plane of the wheel shouldn't change its rate of rotation > wrt an inertial frame if there is no friction. > > If it's initially non-rotating wrt the distant stars (i.e. rotating once per > day wrt the Earth) it will remain so, you can verify that with a bicycle, lift > the front you'll see that the front wheel keeps the same orientation. > > If it's initially non-rotating wrt the Earth, i.e. rotating once per day wrt > the distant stars, it will keep its angular speed i.e. it will remain at rest > wrt the Earth. > > Michel > > - Original Message - > From: "Harry Veeder" <[EMAIL PROTECTED]> > To: > Sent: Friday, July 20, 2007 5:23 AM > Subject: Re: [Vo]:centripetal force question > > >> >> Imagine a ferris wheel (absent the carriages for simplicity) which is >> initially at rest. >> >> As the world turns, the ferris wheel will complete one revolution in a day >> (assuming no friction) with respect to an observer standing >> beside it. >> >> Yes? No? ...or? >> >> Harry >> >> >> >> >> On 15/7/2007 5:17 PM, Michel Jullian wrote: >> >>> Hi Thomas, >>> >>> The (fictitious, or apparent) force you're talking about is a function of >>> _rotations_ (not revolutions) per second, and also of your mass and of your >>> distance from the axis >>> (force=mass*(2*pi*rotations_per_second)^2*radius_of_the_earth assuming you >>> stand on the equator), nothing to do with distance traveled by the planet, >>> and >>> it is not centripetal (going towards the center) but centrifugal (think of >>> fugitive = going away from the center), if it was centripetal it would not >>> subtract from but add to actual weight, which is the actual centripetal >>> force. >>> >>> Michel >>> >>> - Original Message - >>> From: "thomas malloy" <[EMAIL PROTECTED]> >>> To: >>> Sent: Sunday, July 15, 2007 10:02 PM >>> Subject: [Vo]:centripital force question >>> >>> I'm subject to weight loss produced by the centripital force produced by the earth's rotation, I'm wondering if centripetal force is a function of revolutions per time unit, or total distance traveled as the planet travels? --- http://USFamily.Net/dialup.html - $8.25/mo! -- http://www.usfamily.net/dsl.html - $19.99/mo! --- >>> >> >
[Vo]:Re: centripetal force question
Lets' assume the wheel axis is parallel to the Earth's for simplicity. Moving the hub around in the plane of the wheel shouldn't change its rate of rotation wrt an inertial frame if there is no friction. If it's initially non-rotating wrt the distant stars (i.e. rotating once per day wrt the Earth) it will remain so, you can verify that with a bicycle, lift the front you'll see that the front wheel keeps the same orientation. If it's initially non-rotating wrt the Earth, i.e. rotating once per day wrt the distant stars, it will keep its angular speed i.e. it will remain at rest wrt the Earth. Michel - Original Message - From: "Harry Veeder" <[EMAIL PROTECTED]> To: Sent: Friday, July 20, 2007 5:23 AM Subject: Re: [Vo]:centripetal force question > > Imagine a ferris wheel (absent the carriages for simplicity) which is > initially at rest. > > As the world turns, the ferris wheel will complete one revolution in a day > (assuming no friction) with respect to an observer standing > beside it. > > Yes? No? ...or? > > Harry > > > > > On 15/7/2007 5:17 PM, Michel Jullian wrote: > >> Hi Thomas, >> >> The (fictitious, or apparent) force you're talking about is a function of >> _rotations_ (not revolutions) per second, and also of your mass and of your >> distance from the axis >> (force=mass*(2*pi*rotations_per_second)^2*radius_of_the_earth assuming you >> stand on the equator), nothing to do with distance traveled by the planet, >> and >> it is not centripetal (going towards the center) but centrifugal (think of >> fugitive = going away from the center), if it was centripetal it would not >> subtract from but add to actual weight, which is the actual centripetal >> force. >> >> Michel >> >> - Original Message - >> From: "thomas malloy" <[EMAIL PROTECTED]> >> To: >> Sent: Sunday, July 15, 2007 10:02 PM >> Subject: [Vo]:centripital force question >> >> >>> I'm subject to weight loss produced by the centripital force produced by >>> the earth's rotation, I'm wondering if centripetal force is a function >>> of revolutions per time unit, or total distance traveled as the planet >>> travels? >>> >>> >>> --- http://USFamily.Net/dialup.html - $8.25/mo! -- >>> http://www.usfamily.net/dsl.html - $19.99/mo! --- >>> >> >
